Problem: Find $\lim_{x\to -2}\dfrac{3-\sqrt{6x+21}}{x+2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $-2$ (Choice C) C $-3$ (Choice D) D The limit doesn't exist
Substituting $x=-2$ into $\dfrac{3-\sqrt{6x+21}}{x+2}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{3-\sqrt{6x+21}}{x+2} \\\\ &=\dfrac{3-\sqrt{6x+21}}{x+2}\cdot\dfrac{3+\sqrt{6x+21}}{3+\sqrt{6x+21}} \gray{\text{Rationalize the numerator}} \\\\ &=\dfrac{3^2-(6x+21)}{(x+2)(3+\sqrt{6x+21})} \\\\ &=\dfrac{-6\cancel{(x+2)}}{\cancel{(x+2)}(3+\sqrt{6x+21})} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{-6}{3+\sqrt{6x+21}} \text{, for }x\neq -2 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-2$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{3-\sqrt{6x+21}}{x+2}=\dfrac{-6}{3+\sqrt{6x+21}}$ for all $x$ -values in the interval $(-2.5,-1.5)$ except for $x=-2$. Therefore, $\lim_{x\to -2}\dfrac{3-\sqrt{6x+21}}{x+2}=\lim_{x\to -2}\dfrac{-6}{3+\sqrt{6x+21}}=-1$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -2}\dfrac{3-\sqrt{6x+21}}{x+2}=-1$.